Finally I figured out the correct explanation for option (D).
since it is told that the max heap will be constructed by using the insert(insert to the leaf node and then heapify) operation “n” times. Hence, the first key is inserted as no need to heapify with one node therefore no operation, now second key inserted then there are two keys and then we will apply heapify on the second key T.C. O(log n) i.e log 2 (since two keys). Now in this way if we proceed.
The total time complexity will be: "log 2 + log 3 + ….. + log n” which is Summation of log n (n from 2 to n) equates to O(n logn).
A,B,C options are easy and the solution is provided by other user you can refer to that.